Testing for One-Factor Models versus Stochastic Volatility Models



-Xi∕nn}2n ʃj/n^/ (Xs  Xj/n) σ(Xs)dW1 ,s

r- L(n-1) rj / yn-1 1 rl
y
m ^     j = 1 {x3∕n

n ⅛ I

i =1    

<-


Уn-1 1

2=3=1 1 {Xj∕n-xi∕n<M

S/

Gn,m,r

L ( n-1) rj   yn-11

j = 1 {lxj∕n


-Xinn} 2 n ʃj/n )   ( Xs   Xj/n ) μ ( Xs )d s

∑n-1 1

4.


j =1 1 {Xj∕n-Xi∕nn}

SZ

Hn,m,r

L(n-1)rj / yn-11

+ m ^ j = 1 {\Xj/n

n ⅛ I

г=1

S-


-X,,nn}nj/"11 /n (σ2(Xs) - σ2(Xi/n)) ds

n-1

j =1 1 {Xj∕n-Xi∕nn}

--------S/----------------------------

Dn,m,r

Thus, we need to show that Gn,m,r, Hn,m,r and Dn,m,r are oP(1).
Now, because of Lemma 1,

Dn,m,r ≤ √m   sup    Iσ2(Xs) - σ2(Xτ)

Xs-Xτ ≤ξn

≤ √m sup I2(Xτ)   sup   Xs — Xτ

τ [0,1]                Xs-Xτ ≤ξn

= O (√m) Oa.s. (nε/2)Oa.s(ξn) = oa.s. (1),

(15)


(16)


provided that m1 /2nε/2ξn 0. Since m = o(n1 ε), then

Oa.s ( √mnε/2ξn) = 0a.s (n1 /2ξn),

which approaches zero almost surely.

As for Gn,m,r, by the proof of Step 1 of Theorem 1, part(i)a, below, (√n∕√m)Gn,m,r = Gn,r
converges in distribution and so it’s Op(1); therefore Gn,m,r = oP(1), given that m∕n → 0, as
m,n → ∞.

Finally, given the continuity of μ( ),

H[snpι,r ≤ √m sup μ(Xs)    sup    1 Xs — Xi/n1

s [0,1]            |i/n-s 1 /n

s[0,1]

= √mOasi( (nε/4)Oa.s. (n-1 /2 log n) = oa.s. (1).                  (17)

In fact, because of the modulus of continuity of a diffusion (see McKean, 1969, p.96),

sup    1 Xs — Xi/n 1 = Oa.s. n-~1 /2 logn) ,

[i/n-s^i/n
s
[0,1]

and n1 /2+ε/4n-1 /2 log n = n-3ε/4 log n → 0. Therefore, the statement follows.                 

We can now prove Theorem 1.

19



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